TITLE: Matrices with a more general indexing than by integers
QUESTION [6 upvotes]: A matrix is usually understood as an rectangular array of objects (for the most part numbers) arranged in rows and columns. If the objects belong to a set $\mathcal S$, then one can write $M(m,n;\mathcal S)$ for the set of $(m \times n)$-matrices $A$ with entries in $\mathcal S$. Such a matrix $A$ has $m$ rows and $n$ columns. The entry of $A$ occurring in row $i$ and column $j$ is often denoted as $a_{ij}$ and one writes $A = (a_{ij})$.
In linear algebra matrices are used to represent linear maps $f : V \to W$ between finite-dimensional vector spaces $V,W$ with respect to bases $\mathfrak V  =\{v_1,\ldots, v_n\}$ of $V$ and $\mathfrak W  =\{w_1,\ldots, w_m\}$ of $W$. We have $f(v_j) = \sum_{i=1}^m a_{ij}w_i$ with unique $a_{ij}$ and therefore $f(\sum_{j=1}^n\lambda_jv_j) =  \sum_{j=1}^n \lambda_jf(v_j) = \sum_{i,j} \lambda_j a_{ij}w_i$. The $(m\times n)$-matrix $(a_{ij})$ is the matrix representation of $f$ with respect to $\mathfrak V, \mathfrak W$.
For such matrix representations we work only with bases indexed by sets of the special form $I_k = \{1,\ldots,k\}$. Here is my question:

Wouldn't it be more flexible to allow arbitrary finite index sets? That is, to consider matrix sets of the form $M(I, J; \mathcal S) = \mathcal S^{I \times J}$ with arbitrary finite sets $I, J$. A matrix $A \in M(I, J; \mathcal S)$ is then an indexed collection $A 
  =(a_{(i,j)}) \in \mathcal S^{I \times J}$. We can still regard it as a rectangular array of objects of $\mathcal S$ arranged in rows and columns, although these do not have integer row numbers and column numbers.
Of course this concept is not a big innovation. But does is occur somewhere in the literature?

This question was motivated by the answers to  Chain rule for differentiation yields conflicting dimensions. The question deals with a matrix valued function $f : \mathbb R \to M(n,n;\mathbb R)$. Clearly $M(n,n;\mathbb R)$ is isomorphic to $\mathbb R^{n^2}$, but it does not have a canonical basis indexed by $\{1\ldots,n^2\}$. Instead it has a natural base consisting of the matrices $(E_{ij}) \in M(n,n;\mathbb R)$ where the $E_{ij}$ have an entry $1$ in row $i$ and column $j$, all other entries being $0$. The index set of this natural basis is $I_{m,n} = \{1,\ldots,m\} \times \{1,\ldots,n\}$. When considering the derivative $Df \mid_x$ at $x \in \mathbb R$, which naturally is a linear map $\mathbb R \to M(n,n;\mathbb R)$ , the mentioned answers are wavering around by saying that we have to "flatten matrices" or to "identify $M(n,n;\mathbb R)$ with $\mathbb R^{n^2}$" to get a matrix in $M(n^2,1;\mathbb R)$. One can do this, but I think it is uncessary and may even cause confusion (the OP of the above question seems to mix up $M(n^2,1;\mathbb R)$ with $M(n,n;\mathbb R)$ when flattening $Df\mid_x)$). In my opinion it is much more transparent to say that the Jacobian of $f$ at $x$ is a matrix in $M(I_{m,n},1;\mathbb R)$.

REPLY [2 votes]: I decided to write an extended comment in form of a community wiki to Andrew D. Hwang's  excellent answer. I only consider the finite-dimensional case although there are analogues for infinite dimension.
The "indexing-problem" arises already on the level of representing vectors of a finite-dimensional vector space $V$ via their coordinates with respect to a basis $\mathfrak V$ of $V$. A basis of $V$ is usually defined as a subset $\mathfrak V \subset V$ which is linearly independent and spans $V$. In most definitions $\mathfrak V$ is written in the form $\{v_1,\ldots,v_n\}$. This implicitly says that we actually consider ordered bases which are finite ordered sequences $(v_1,\ldots,v_n)$ rather than finite (unordered) sets. In other words, we choose a bijecton $I_n = \{1,\ldots,n\} \to \mathfrak V$ so that the integers $i = 1,\ldots,n$ are used to index the elements of $\mathfrak V$. This seems to be quite natural because explicitly writing down the elements of $\mathfrak V$ is usually done in form of a list, i.e. in form of a linear sequential arrangement (either from left to right  or top-down).
In my opinion working with ordered bases is conceptually unneccesary for most purposes of linear algebra. An exception is the concept of orientation. In fact, each element $\mathbf x \in V$ can be written as $\mathbf x = \sum_{v \in \mathfrak V} x_v v$ with unique $x_v \in K$. In other words, when working with a basis $\mathfrak V$ we can avoid the identification $V \approx K^n$ and use the canonical isomorphism $V \approx K^{\mathfrak V}$. That is, the  coordinate representation of $\mathbf x$ with respect to $\mathfrak V$ is written abstractly as $(x_v)_{v \in \mathfrak V}$ and not as $(x_i)_{i \in I_n}$.
The problem here is that if we want to explicitly write down $(x_v)_{v \in \mathfrak V}$ for concrete numerical values of the $x_v$, then we have to agree upon a definite arrangement pattern for the $x_v$ on a sheet of paper which allows to identify the index $v$ from the position of the element $x_v \in K$ in the arrangement pattern. The usual way is of course to do it in form of a linear sequential arrangement either from left to right, giving row vectors,  or top-down, giving column vectors. This is nothing else than the choice of a bijection $I_n \to  \mathfrak V$. However, we could also agree to arrange the $x_v$ at $n$ positions of a circle (which requires to specify a function from $\mathfrak V$ to the circle). In some cases we even have a non-linear arrangement pattern by definition of $V$. This is the case for $V  =M(m,n;K)$. This vector space has a canonical basis in form of the $E_{ij}$ (see the question). The coordinate representation of a matrix $A $ with respect to this basis is naturally written in matrix pattern which gives us again $A$. Of course we can flatten it, but this would be unnatural and arbitrary.